Problem: Simplify; express your answer in exponential form. Assume $r\neq 0, t\neq 0$. $\dfrac{{(r^{3})^{-5}}}{{(r^{-5}t)^{-4}}}$
To start, try working on the numerator and the denominator independently. In the numerator, we have ${r^{3}}$ to the exponent ${-5}$ . Now ${3 \times -5 = -15}$ , so ${(r^{3})^{-5} = r^{-15}}$ In the denominator, we can use the distributive property of exponents. ${(r^{-5}t)^{-4} = (r^{-5})^{-4}(t)^{-4}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(r^{3})^{-5}}}{{(r^{-5}t)^{-4}}} = \dfrac{{r^{-15}}}{{r^{20}t^{-4}}}$ Break up the equation by variable and simplify. $\dfrac{{r^{-15}}}{{r^{20}t^{-4}}} = \dfrac{{r^{-15}}}{{r^{20}}} \cdot \dfrac{{1}}{{t^{-4}}} = r^{{-15} - {20}} \cdot t^{- {(-4)}} = r^{-35}t^{4}$.